If we draw a card from a standard 52-card deck, what is the probability that the card is neither a heart nor a queen?

So here we are being asked to draw a card that isn't a heart or a queen... the amount of cards that are not hearts or queens out of a 52 card deck is 36. So that makes the probability 36/52 if we want to make that a percent then we have to change the denominator to a 100 we end up with this problem:

36/52 = x/100

There are many ways we can solve this. One way being cross multiplying, which I think is the easiest.

If our problem is a/b = c/d we multiply a and d then we multiply b and c

Plugging our original problem into that, we 36 times 100 and 52 times x. That gives us 3600 and 52x

This is now an algebraic problem.

52x=3600

First step, divide each side by 52

We get x=69.2 rounded to the nearest tenth.

So our probability is about 69/100 or 69%

Any Questions?

proability is (desired outcomes)/(total possible outcomes)

total possible outcomes is 52 cards

out of those 52 cards, how many are NOT heart or queen?

hearts is a suit, so 1/4 of the deck (or 52/4=13 cards) are not what we want

queens, there are 4 queens in a deck, but one of them is a queen of hearts. to avoid doublecount, say 3 queens

so total of 13+3=16 cards to not pick

so we have 52-16=36 cards we want to pick

probablity=36/52=18/26=9/13≈63.23%

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.